A 'pop' gun consists of a tube 25 cm long closed at one end by a cork and at the other end by a tightly fitted piston. The piston is pushed slowly in. When the pressure rises to one and half times the atmospheric pressure, the cork is violently blown out. Calculate the frequency of the 'pop' caused by its ejection. Speed ofsound in air is 340 m/s

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
We have a pop gun consisting of a tube that is 25 cm long, closed at one end by a cork and at the other end by a tightly fitted piston. The piston is pushed in until the pressure reaches 1.5 times atmospheric pressure, causing the cork to be blown out.

### Step 2: Identify the type of wave in the tube
Since one end of the tube is closed (the cork) and the other end is open (the piston), the tube behaves like a closed organ pipe. In a closed organ pipe, there is a node at the closed end and an antinode at the open end.

### Step 3: Determine the wavelength
For a closed organ pipe, the fundamental frequency corresponds to a quarter of the wavelength fitting in the length of the tube:
\[ \text{Length of the tube} = L = 25 \text{ cm} = 0.25 \text{ m} \]
The relationship between the wavelength (\(\lambda\)) and the length of the tube is given by:
\[ L = \frac{\lambda}{4} \implies \lambda = 4L = 4 \times 0.25 \text{ m} = 1 \text{ m} \]

### Step 4: Use the speed of sound to find the frequency
The frequency (\(f\)) of the sound produced can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
where \(v\) is the speed of sound in air, given as 340 m/s. Substituting the values:
\[ f = \frac{340 \text{ m/s}}{1 \text{ m}} = 340 \text{ Hz} \]

### Step 5: Calculate the effective length after pressure change
When the pressure rises to 1.5 times atmospheric pressure, we can use Boyle's law, which states that \(P_1 L_1 = P_2 L_2\). Here:
- \(P_1 = P_0\) (atmospheric pressure)
- \(L_1 = 0.25 \text{ m}\)
- \(P_2 = 1.5 P_0\)
- \(L_2\) is the new length we need to find.

Using Boyle's law:
\[ P_0 \cdot 0.25 = 1.5 P_0 \cdot L_2 \]
Dividing both sides by \(P_0\):
\[ 0.25 = 1.5 \cdot L_2 \implies L_2 = \frac{0.25}{1.5} \approx 0.1667 \text{ m} \]

### Step 6: Calculate the new frequency
Now, we can find the new frequency using the new length:
\[ \lambda = 4L_2 = 4 \times 0.1667 \text{ m} \approx 0.6667 \text{ m} \]
Now substituting back into the frequency formula:
\[ f = \frac{340 \text{ m/s}}{0.6667 \text{ m}} \approx 510 \text{ Hz} \]

### Final Answer
The frequency of the 'pop' caused by the ejection of the cork is approximately **510 Hz**.

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